Given a non-zero complex number $z=a+ib$, I want to show that there exists a unique $\theta\in (-\pi,\pi]$, satisfying the equations $$a=|z|\cos \theta,\mbox{and } b=|z|\sin \theta.$$ Now, since $|a|\leq|z|$, we deduce the inequality
$$-1\leq\dfrac{a}{|z|}\leq 1.$$ Also as $\cos:[0,\pi]\to [-1,1]$ is bijective, there exist a unique $\theta\in[0,\pi]$ such that $\cos \theta =\frac{a}{|z|} \implies a=|z|\cos \theta.$ Next I will show that for this same $\theta$, the equation $b=|z|\sin \theta$ is also satisfied. We have, $$|z|\sin \theta=\sqrt{(a^2+b^2)\sin^2 \theta} =\sqrt{(a^2+b^2)(1-\cos^2 \theta)}=\sqrt{(a^2+b^2)-(a^2+b^2)\cos^2\theta}\\ =\sqrt{(a^2+b^2)-|z|^2\cos^2\theta}=\sqrt{(a^2+b^2)-a^2}\\ =\sqrt{b^2}=b.$$
Thus we have got a unique $\theta\in [0,\pi]$ satisfying the above two equations but NOT in $(-\pi,\pi]$. What is wrong with my calculations? what did I miss? In many complex analysis books, I have seen this $\theta$ is nothing but the principal argument of $z$, and it should lie in the interval $(-\pi,\pi]$. Please help anyone I'm confused.
If $z=|z|(cosθ+isinθ)$, then $cosθ=\frac {x}{|z|}$ and $sinθ=\frac {y}{|z|}, z\neq 0$ (1). If $θ\in (-π,π]$, then there is a unique $θ$ that satisfies both equations (1), because the functions sine and cosine have period $2π$, so there is no $T\in \Bbb R$, such that $θ+T\in (-π,π]$ and $cosθ=cos(θ+Τ)$ or $sinθ=sin(θ+Τ)$.
The answer about existence of such a $θ$ is trivial. Let a complex number $z=x+yi, x,y\in \Bbb R$.We define $|z|^2:=x^2+y^2$, so there exists a $θ$, such that $x=|z|cosθ, y=|z|sinθ$, for $x^2+y^2=(|z|cosθ)^2+(|z|sinθ)^2=|z|^2(cos^2θ+sin^2θ)=|z|^2$.